International Journal of Pure and Applied Volume 75 No. 1 2012, 19-68 AP ISSN: 1311-8080 (printed version) url: http://www.ijpam.eu ijpam.eu

A CERTAIN p-ADIC SPECTRAL THEOREM

R.L. Baker Department of Mathematics University of Iowa Iowa City, Iowa, 52242, USA

Abstract: We extend a p-adic spectral theorem of M. M. Vishik to a certain class of p-adic Banach algebras. This class includes inductive limits of finite- dimensional p-adic Banach algebras of the form B(X ), where X is a p-adic Banach space of the form X ≃ Ωp(J), J being a finite nonempty set. In particular, we present a p-adic spectral theorem for p-adic UHF algebras and p-adic TUHF algebras (Triangular UHF Algebras).

AMS Subject Classification: 12J99, 46L99, 46S10 Key Words: p-adic Banach algebras, p-adic UHF algebras, p-adic spectral theorem, p-adic triangular UHF algebras

1. Introduction

In an germinal 1960 paper [6] J. Glimm introduced the class of uniformly hyper- finite (UHF) C∗-algebras. In that paper Glimm showed that UHF algebras are classified up to ∗-isomorphisms by their supernatural . Glimm’s classi- fication of UHF C∗-algebras inspired the author of the present paper to intro- duce in [2] a class of non-selfadjoint operator algebras that are currently called standard triangular UHF (TUHF) algebras. Briefly: a standard triangular UHF (TUHF) operator algebra is any unital Banach algebra that is isometrically iso- morphic to a Banach algebra inductive limit of the form T = lim(Tp ; σp p ); −→ n n m where (pn) is a sequence of positive such that pm | pn whenever m ≤ n.

c 2012 Academic Publications, Ltd. Received: November 11, 2011 url: www.acadpubl.eu 20 R.L. Baker

Here Tpn is the algebra of pn × pn upper triangular complex matrices and for m ≤ n, σpn pm : Tpm → Tpn is the mapping x ֒→ 1 ⊗ x = diag(x,...,x). The main result in [2] is that two standard triangular UHF operator algebras are iso- metrically isomorphic if, and only if they have the same supernatural . This result can be extended in at least three different directions. One direction is to view standard triangular UHF operator algebras as special cases of Banach algebra inductive limits of upper triangular matrix algebras, and the the goal is to use purely Banach-algebraic methods to extend the main result in [2] to these inductive limits. Such an extension is presented in the paper [3], where it is proved that the supernatural number associated to an arbitrary triangular UHF (TUHF) Banach algebra is an isomorphism invariant of the algebra, pro- vided that the algebra satisfies certain “local dimensionality conditions.” The proof of main result of [3], although relying only on classical complex Banach algebra techniques, makes essential use of the classical spectral theorem (the Riesz functional calculus) for complex Banach algebras. A second direction in which the main result in [2] has been extended is to view standard TUHF algebras as triangular subalgebras of UHF C∗-algebras, such that the diagonal of the triangular subalgebra is a canonical masa in the ambient C∗-algebra. In this vein, the results in [2] have been substantially extended by J.R. Peters, Y.T. Poon and B.H. Wagner [11], and by S.C. Power [9], [10]. Finally, a third direction in which the principal results of [2] can be extended is to convert these results into a “pure piece” of p-adic functional analysis (to borrow a turn of phrase used by I. Kaplansky). For any p, let Ωp be the p-adic counterpart of the complex numbers C (see [7], page 13). Replacing C by Ωp in the definition of complex (T)UHF Banach algebras, we obtain the definition of p-adic (T)UHF Banach algebras. The results in [2] for complex TUHF algebras can be duplicated for p-adic TUHF algebras, provided that a sufficiently gen- eral p-adic version of the Riesz functional calculus can be developed for p-adic Banach algebras ([3]). The purpose of the present paper it to prove a p-adic spectral theorem (Theorem 2.10) that is an extension of a classical p-adic spec- tral theorem of M. M. Vishik (see [7], page 149). We then use this extended p-adic spectral theorem to prove a p-adic Spectral Theorem for p-adic TUHF and UHF Banach algebras (Theorem 3.3). In the preprint [4] we use Theorem 3.3, Theorem 3.4, and Lemma 3.5 to transfer, mutatis mutandis, the results of [3] to p-adic TUHF Banach algebras. In the remainder of the present section of the paper we put forth the prelim- inary material on p-adic analysis and p-adic Banach algebras that is necessary for proving Theorem 2.10. Then in Section 2 we prove Theorem 2.10. Finally, in Section 3, we use Theorem 2.10 to prove Theorem 3.3. A CERTAIN p-ADICSPECTRALTHEOREM 21

Let p be a prime number, and let Q be the field of rational numbers. Let | · |p be the function defined on Q by

a = pordpb−ordpa, |0| = 0. b p p

Here ordp of a non-zero is the highest power of p dividing the integer. Then | · |p is a norm on Q. The field Qp is defined to be the completion of Q under the norm | · |p. Define Qp to be the algebraic closure of Qp (see [7]: p. 11). Unlike the case of the real numbers R, whose algebraic closure C is only a quadratic extension of R, the algebraic closure Qp of Qp has infinite degree over Qp. However, the norm | · |p on Qp can be extended to a norm | · |p on Qp. But it turns out that Qp is not complete under this extended norm. Thus, in order to do analysis, we must take a larger field than Qp. We denote the completion of Qp under the norm | · |p by Ωp, that is,

∧ Ωp =Qp,

∧ where means completion with respect to | · |p. The symbol “C” is usually used to denote Ωp (see [7], page 13 and [12], page 140).

Definition 1.1. Let r ≥ 0 be an non-negative , and let a ∈ Ωp and let σ ⊆ Ωp. We have the following definitions.

Da(r)= { x ∈ Ωp | |x − a|p ≤ r }; − Da(r )= { x ∈ Ωp | |x − a|p < r };

Dσ(r)= { x ∈ Ωp | dist(x,σ) ≤ r }; − Dσ(r )= { x ∈ Ωp | dist(x,σ) < r }.

− − − If b ∈ Da(r), then Db(r) = Da(r), and if b ∈ Da(r ), then Db(r ) = Da(r ). − Thus, any point in a disc is its center. Hence Da(r), Da(r ) are both open and closed in the topological sense. It is conventional to take inf ∅ =+∞, hence for − r> 0 we have D∅(r )= ∅ and D∅(r)= ∅ (see [5], page 4).

Lemma 1.2. Let ∅= 6 σ ⊆ Ωp be compact. Then for any s> 0 there exists 0 < r ∈ |Ωp|p and a1, ..., aN ∈ σ such that r

N

Dσ(r)= Dai (r); i [=1 22 R.L. Baker

N − σ ⊆ Dai (r ), i [=1 where the Dai (r) are disjoint.

Proof. Since σ is compact, there exist a1, ..., aN ∈ σ such that

N − σ ⊆ Dai (s ), (1) i [=1 − where the union is disjoint and for 1 ≤ i ≤ N, σ ∩ Dai (s ) 6= ∅. Now, for each − − 1 ≤ i ≤ N, Dai (s ) is closed, and hence σ ∩ Dai (s ) is compact. Therefore − there exists b1, ..., bN ∈ σ such that for 1 ≤ i ≤ N, bi ∈ σ ∩ Dai (s ) and

− |bi − ai|p = sup{ |x − ai|p : x ∈ σ ∩ Dai (s ) }.

− Because each bi ∈ Dai (s ), we have, for 1 ≤ i ≤ N, s > |bi − ai|p. Now select any r ∈ |Ωp|p such that r> 0 and

s>r> |b1 − a1|p,..., |bN − aN |p.

We claim that N − σ ⊆ Dai (r ), (2) i [=1 where Da1 (r),...,DaN (r) are disjoint. To prove the claim, let a ∈ σ; then by − − (1), a ∈ Dai (s ) for some 1 ≤ i ≤ N, and hence a ∈ σ ∩ Dai (s ), which implies that

− |a − ai|p ≤ sup{ |x − ai|p : x ∈ σ ∩ Dai (s ) } = |bi − ai|p < r.

− This shows that a ∈ Dai (r ), which proves (2). Now let 1 ≤ i, j ≤ N, with i 6= j. Suppose that Dai (r) ∩ Daj (r) 6= ∅, and let a ∈ Dai (r) ∩ Daj (r). Then we have |ai − aj|p ≤ max{ |ai − a|p, |a − aj|p }≤ r

− − hence ai ∈ Dai (s ) ∩ Daj (s ) = ∅. This contradiction proves the claim. We now claim that N

Dσ(r)= Dai (r). (3) i [=1 A CERTAIN p-ADICSPECTRALTHEOREM 23

N

It is clear that Dai (r) ⊆ Dσ(r). Let x ∈ Dσ(r), then because σ is compact, i=1 there exists a ∈Sσ such that r ≥ dist(x,σ)= |x − a|p. Hence, x ∈ Da(r). Now, − a ∈ σ, so by (2), there exists 1 ≤ j ≤ N such that a ∈ Daj (r ) ⊆ Daj (r). N

Therefore, Da(r) = Daj (r), which shows that x ∈ Dai (r). This proves i=1 (3). S

Lemma 1.3. Let ∅= 6 σ ⊆ Ωp. Let r> 0 and let b1, . . . , bM be in Ωp, with

M − σ ⊆ Dbi (r ), (1) i [=1 where the Dbi (r) are disjoint. Then there exist a1, . . . , aN in σ and ∅= 6 I ⊆

{ 1,...,M } such that the Dai (r) are disjoint and

N − − − Dσ(r )= Dai (r )= Dbi (r ); (2) i i I [=1 [∈ N

Dσ(r)= Dai (r)= Dbi (r). i i I [=1 [∈

Proof. Let ∅= 6 I ⊆ { 1,...,M } be the subset of { 1,...,M } consisting of − those 1 ≤ i ≤ M for which σ ∩ Dbi (r ) 6= ∅. Write

I = { k1,...,kN }.

For each 1 ≤ i ≤ N, select an a ∈ σ ∩ D (r−). Then for 1 ≤ i ≤ N, we have i bki D (r−)= D (r−), and hence, by (1), ai bki

N − − σ ⊆ Dai (r )= Dbi (r ). (3) i i I [=1 [∈

The assumption that the Dbi (r) are disjoint easily implies that the Dai (r) are disjoint. We claim that

N − − Dσ(r ) ⊆ Dai (r ); (4) i [=1 24 R.L. Baker

N

Dσ(r) ⊆ Dai (r). i [=1 − To prove (4) let x ∈ Dσ(r ). Then, by definition, dist(x,σ) < r, hence there exists a ∈ σ such that |x − a|p < r. Then (3) implies that there exist 1 ≤ j ≤ N such that |a − aj|p < r. Therefore, we have

|x − aj|p = |(x − a) + (a − aj)|p ≤ max{ |x − a|p, |a − aj|p } < r.

− Thus, x ∈ Daj (r ). This proves the first part (4). To prove the second part of N

(4), suppose that there exists x ∈ Dσ(r) with x 6∈ Dai (r). Then |x−ai|p > r i=1 for 1 ≤ i ≤ N, hence we can find 0 s for 1 ≤ i ≤ N. Now let a ∈ σ be arbitrary, then by (3), |a − ai|p < r for some 1 ≤ i ≤ N. We then have |x − a|p = |(x − ai) + (ai − a)|p = |x − ai|p > s. (5)

In (5) we have used the easily proved fact that for y, z ∈ Ωp, if |y|p > |z|p, then |y + z|p = |y|p (see [?], Theorem 2, page 5). But a ∈ σ is arbitrary, hence (5) − implies that dist(x,σ) ≥ s > r, which contradicts x ∈ Dσ(r ). This proves the second part of (4). It is easy to see that

N − − Dai (r ) ⊆ Dσ(r ); (6) i [=1 N

Dai (r) ⊆ Dσ(r). i [=1 Hence by (4) and (6), we see that (2) holds.

Lemma 1.4. Let a ∈ Ωp and 0 < r ∈ |Ωp|p. Suppose that f : Da(r) → Ωp is the uniform limit of rational functions with poles outside Da(r). Then there k exists a sequence (ck) in Ωp such that lim r |ck|p = 0 and for all x ∈ Da(r) k→∞

∞ k f(x)= ck(x − a) . Xk=0 Proof. See [7], Lemma 3, page 130.

Lemma 1.5. Let a ∈ Ωp, and let 0 < r ∈ |Ωp|p. Let f : Da(r) → Ωp. k Suppose that there exists a sequence (ck) in Ωp be such that lim r |ck|p = 0, k→∞ A CERTAIN p-ADICSPECTRALTHEOREM 25

and for all x ∈ Da(r), ∞ k f(x)= ck(x − a) . Xk=0 Then max |f(x)|p is attained when |x − a|p = r, and we have x∈Da(r)

k max |f(x)|p = max r |ck|p . x∈Da(r) k

Proof. See [7], Lemma 3, page 130.

Definition 1.6. Let a ∈ Ωp, and let 0 < r ∈ |Ωp|p. Let ∅= 6 σ ⊆ Ωp. A function f : Da(r) → Ωp is said to be Krasner analytic on Da(r) iff f can be ∞ k represented by a power series on Da(r) of the form f(x) = k=0 ck(x − a) , k where lim r |ck|p = 0. Define Br(σ) to be the set of all functions k→∞ P

f : Dσ(r) → Ωp such that f is Krasner analytic on Da(r) whenever a ∈ Ωp and Da(r) ⊆ Dσ(r). If σ is compact, we define L(σ) = { Br(σ) | 0 < r ∈ |Ωp|p }, and we call L(σ) the set of locally analytic functions on σ (see [7], page 136). S Definition 1.7. Let ∅= 6 σ ⊆ Ωp. For 0 < r,s ∈ |Ωp|p and a1, . . . , aN , b1,..., bM ∈ Ωp, define the order relation (s, b1, . . . , bM ) ≤ (r, a1, . . . , aN ) by

M N

(s, b1, . . . , bM ) ≤ (r, a1, . . . , aN ) ⇐⇒ s ≤ r and Dbi (s) ⊆ Dai (r). i i [=1 [=1 For 0 < r ∈ |Ωp|p and a1, . . . , aN ∈ Ωp, define

N

Br,a1,...,aN = f : Dai (r) → Ωp f is Krasner analytic on each Dai (r) .  i=1  [

Let I(σ) to be the set of all (r, a1, . . . , aN ) such that:

(1) 0 < r ∈ |Ωp|p and a1, . . . , aN ∈ Ωp;

(2) Dai (r) ∩ Daj (r)= ∅ for i 6= j. N − (3) σ ⊆ Dai (r ). i=1 Finally, defineS

L(σ)= { Br,a1,...,aN | (r, a1, . . . , aN ) ∈ I(σ) }. [ 26 R.L. Baker

Lemma 1.8. Let ∅= 6 σ ⊆ Ωp. Then the following statements hold.

(a) The set I(σ) is decreasingly filtered under the order relation ≤.

(b) Let (r, a1, . . . , aN ) ∈ I(σ) and f ∈ Br,a1,...,aN . Then f is bounded on N

U = Dai (r), and we define the uniform norm kfku of f by i=1 S kfku = max |f(x)|p. x∈U

(c) Let (s, b1, . . . , bM ), (r, a1, . . . , aN ) ∈ I(σ), with

(s, b1, . . . , bN ) ≤ (r, a1, . . . , aN ).

M

Define V = Dbi (s), then f|V ∈ Bs,b1,...,bM and the mapping f 7→ f|V i=1 is continuousS on Br,a1,...,aN in the uniform norm.

(d) L(σ) is an algebra over Ωp.

(e) For α = (r, a1, . . . , aN ) ∈ I(σ), set Bα = Br,a1,...,aN . For β = (s, b1, . . . , bM ) M α in I(σ), with β ≤ α, let V = Dbi (s) and define ϕβ : Bα → Bβ by i=1 α S ϕβ (f)= f|V , all f ∈ Bα.

Then for α ≥ β ≥ γ ∈ I(σ) following conditions are satisfied

α ϕα(f)= f, for all f ∈ Bα, α β α ϕβ ϕγ = ϕγ .

It is worth noting here that we can define the inverse limit lim Bα of the ←− system { Bα | α ∈ I(σ) } and place the projective topology on lim Bα. But ←− we will not make use of this topology in the present paper.

(f) If σ is compact, then L(σ)= L(σ), i.e.,

L(σ)= { Br(σ) | 0 < r ∈ |Ωp|p }. (1) [

(g) For all (r, a1, . . . , aN ) ∈ I(σ), if a1, . . . , aN ∈ σ, then Br,a1,...,aN = Br(σ). A CERTAIN p-ADICSPECTRALTHEOREM 27

Proof. To prove (a) let (s, b1, . . . , bM ), (r, a1, . . . , aN ) be members of I(σ) such that (s, b1, . . . , bM ) ≤ (r, a1, . . . , aN ). Suppose that 1 ≤ i ≤ M, 1 ≤ j ≤ N, and

− − Dbi (s ) ∩ Daj (r ) 6= ∅.

− − Let x ∈ Dbi (s ) ∩ Daj (r ), then |bi − x|p

− − − − − Dbi (s )= Dbi (s ) ∩ Dbi (r )= Dbi (s ) ∩ Daj (r ).

It follows that if I is the set of all 1 ≤ i ≤ M such that there exists a 1 ≤ j ≤ N for which − − Dbi (s ) ∩ Daj (r ) 6= ∅, then we have

M N − − − σ ⊆ Dbi (s ) ∩ Daj (r ) = Dbi (s ). i j i I  [=1   [=1  [∈

Hence if we write (bi)i∈I = (c1,...,cK ), then we have (s, c1,...,cK ) ∈ I(σ), with

(s, c1,...,cK ) ≤ (s, b1, . . . , bM ) and (s, c1,...,cK ) ≤ (r, a1, . . . , aN ).

This proves (a). To prove (b), let 1 ≤ i ≤ N be arbitrary. Because f is Krasner analytic on Dai (r), Lemma 1.5 implies that max |f(x)|p is attained on the circle x∈Dai (r) |x − ai|p = r. It follows that f is bounded on U. N

For the proof of (c), define U = Dai (r). Let f ∈ Br,a1,...,aN and let i=1 1 ≤ i ≤ M. Because (s, b1, . . . , bN ) ≤ S(r, a1, . . . , aN ), we have 0 < s < r and

V ⊆ U. Therefore there exists a 1 ≤ j ≤ N such that bi ∈ Daj (r), and hence

Dbi (r) = Daj (r). Because f is Krasner analytic on Daj (r), it follows from

Lemma 1.4 that f is Krasner analytic on Dbi (r). Hence, because s ≤ r, f is

Krasner analytic on Dbi (s). This shows that f|V ∈ Bs,b1,...,bM . We have V ⊆ U, therefore it is clear that the mapping f 7→ f|V is continuous on Br,a1,...,aN in the uniform norm. 28 R.L. Baker

The proof of (d) follows form (a); and the proof of (e) follows from (c). To prove (f), assume that σ is compact. Let f ∈ L(σ). Then by Definition 1.7 there exist a1, . . . , aN in Ωp and 0 < r ∈ |Ωp|p such that

N − σ ⊆ Dai (r ), (2) i [=1 where the Dai (r) are disjoint and f is Krasner analytic on each Dai (r). Then by (2) and Lemma 1.3, there exists ∅= 6 I ⊆ { 1,...,N } such that

Dσ(r)= Dai (r). (3) i I [∈

Let b ∈ Ωp, with Db(r) ⊆ Dσ(r). Then by (3), there exists i ∈ I such that b ∈ Dai (r), and consequently, Db(r) = Dai (r). Because f is Krasner analytic on Dai (r) we see that f : Db(r) → Ωp is the uniform limit of rational functions with poles outside Db(r). It follows from Lemma 1.4 that f is Krasner analytic on Db(r). This shows that f ∈ Br(σ). Because f ∈L(σ) is arbitrary, we get

L(σ) ⊆ { Br(σ) | 0 < r ∈ |Ωp|p }. (4) [ Now let f ∈ { Br(σ) | 0 < r ∈ |Ωp|p }, say, f ∈ Br(σ), 0 < r ∈ |Ωp|p. By Lemma 1.2 there exist b , . . . , bM in σ and 0 < s ∈ |Ωp|p such that s < r, the S 1 Dbi (s) are disjoint, and

M M − Dσ(s)= Dbi (s), σ ⊆ Dbi (s ). (6) i i [=1 [=1

Let 1 ≤ i ≤ M be arbitrary. Because bi ∈ σ, we have Dbi (r) ⊆ Dσ(r). By assumption, f ∈ Br(σ), therefore f is Krasner analytic on Dbi (r). Hence, by Definition 1.6, f can be represented by a power series on Dbi (r) of the ∞ k k form f(x) = ck(x − bi) , where lim r |ck|p = 0. Since s < r, we have k=0 k→∞ ∞ P k Dbi (s) ⊆ Dbi (r), and hence, for x ∈ Dbi (s), we have f(x) = ck(x − bi) , k=0 k with lim s |ck|p = 0. This shows that f is Krasner analytic onPDbi (s). This k→∞ proves that f ∈ Bs,b1,...,bM . It follows that

{ Br(σ) | 0 < r ∈ |Ωp|p }⊆L(σ). (7) [ A CERTAIN p-ADICSPECTRALTHEOREM 29

Statements (4) and (7) together imply (1). Finally, statement (g) is an easy consequence of Lemma 1.3 and Lemma 1.4.

Definition 1.9. (The Shnirelman Integral) Let 0 < r ∈ |Ωp|p. Let a ∈ Ωp, and let f be an Ωp-valued function whose domain contains all x ∈ Ωp such that | x − a |p = r. Let Γ ∈ Ωp, with |Γ|p = r. Then the Shnirelman integral of f over the circle { x ∈ Ωp : |x − a|p = r } is defined to be the following limit, provided the limit exists.

1 f(x) dx = lim f(a + ξΓ). n→∞ n p6|n ξn a,ZΓ X=1

Definition 1.10. Let ∅= 6 σ ⊆ Ωp be compact, let H0(σ) denote the set of functions ϕ : σ → Ωp which are Krasner analytic on the complement σ of σ, i.e.,

(1) ϕ is the limit of rational functions whose poles are contained in σ, the limit being uniform in any set of the form

− Dσ(r)= { z ∈ Ωp | dist(z,σ) ≥ r }, σ ⊆ Dσ(r ),r> 0.

(2) lim ϕ(z) = 0. |z|p→∞

Lemma 1.11. Let ∅ 6= σ ⊆ Ωp be compact. Let 0 < r1, r2 ∈ Ωp, with r1 ≤ r2. Let Γ1, Γ2 be in Ωp, with |Γ1|p = r1, |Γ2|p = r2. Assume that

a1, . . . , aM ; b1, . . . , bN ∈ Ωp are given, with

M M − Dσ(r1)= Dai (r1), σ ⊆ Dai (r1 ); i i [=1 [=1 N N − Dσ(r2)= Dbi (r2), σ ⊆ Dbi (r2 ), i i [=1 [=1 30 R.L. Baker

where the Dai (r1) and the Dbi (r2) are disjoint. Then for ϕ ∈ H0(σ) and f ∈ Br2 (σ), we have f ∈ Br1 (σ), and the following sums exist and are equal. M N f(x)(x − ai)ϕ(x) dx = f(x)(x − bi)ϕ(x) dx. i=1 Z i=1 Z X ai,Γ1 X bi,Γ2

Proof. See [7], Lemma 8, page 138.

Lemma 1.12. Let ∅= 6 σ ⊆ Ωp be compact. Let 0 < r ∈ |Ωp|p and let Γ ∈ Ωp, with |Γ|p = r. Assume that

M

Dσ(r)= Dbi (r); (1) i [=1 M − σ ⊆ Dbi (r ), i [=1 where b1, . . . , bM ∈ σ and the Dbi (r) are disjoint. Let ϕ ∈ H0(σ). Suppose that a1, . . . , aN ∈ Ωp, with N − σ ⊆ Dai (r ), (2) i [=1 where the Dai (r) are disjoint. Finally, assume that f is Krasner analytic on each of the Dai (r). Then f ∈ Br(σ) and M N f(x)(x − bi)ϕ(x) dx = f(x)(x − ai)ϕ(x) dx. i=1 Z i=1 Z X bi,Γ X ai,Γ

Proof. By (2) and Lemma 1.3, there exists ∅= 6 I ⊆ { 1,...,N } such that

− − Dσ(r )= Dai (r ), Dσ(r)= Dai (r). (3) i I i I [∈ [∈ To show that f ∈ Br(σ), let b ∈ Ωp, with Db(r) ⊆ Dσ(r). Then (3) implies that there exists i ∈ I such that b ∈ Dai (r), and hence Db(r)= Dai (r); because f is Krasner analytic on Dai (r), it follows from Lemma 1.4 that f is Krasner analytic on Db(r). This proves that f ∈ Br(σ). Now, by (1), (3) and Lemma 1.11 we have M f(x)(x − bi)ϕ(x) dx = f(x)(x − ai)ϕ(x) dx. (4) i=1 Z i∈I Z X bi,Γ Xai,Γ A CERTAIN p-ADICSPECTRALTHEOREM 31

Now let 1 ≤ j ≤ N, with j 6∈ I. We claim that Daj (r) ⊆ σ. For suppose that b ∈ σ, with b ∈ Daj (r). Then because σ ⊆ Dσ(r), (3) implies that b ∈ Dai (r) for some i ∈ I. We have i 6= j and b ∈ Dai (r) ∩ Daj (r). This contradicts the assumption that Dai (r) and Daj (r) are disjoint when i 6= j. This proves the claim. By assumption, ϕ ∈ H0(σ), and since f ∈ Br(σ), we see that f(x)(x − aj)ϕ(x) is Krasner analytic on Daj (r) for each 1 ≤ j ≤ N such that j 6∈ I. It follows from the p-adic Cauchy integral formula (see [7], Lemma 4, page 131) that for j 6∈ I, f(x)(x − aj)ϕ(x) dx = 0. Therefore, (4) gives aj ,Γ R M f(x)(x − bi)ϕ(x) dx = f(x)(x − ai)ϕ(x) dx i=1 Z i∈I Z X bi,Γ Xai,Γ

= f(x)(x − ai)ϕ(x) dx+ i∈I Z Xai,Γ

f(x)(x − ai)ϕ(x) dx i6∈I Z Xai,Γ N = f(x)(x − ai)ϕ(x) dx. i=1 Z X ai,Γ This completes the proof of the lemma.

Definition 1.13. A p-adic Banach space over Ωp is a vector space X over Ωp together with a norm k·kp from X to the nonnegative real numbers such that for all x,y ∈X : (a) kxkp = 0 if and only if x = 0; (b) kx+ykp ≤ max{ kxkp, kykp }; (c) kaxkp = |a|pkxkp; (d) X is complete under k · kp. We shall assume that kX kp = |Ωp|p, i.e., for every x 6=0in X there exists a ∈ Ωp such that kaxkp = 1. ∗ The dual X of a p-adic Banach space over Ωp is defined in the usual way. If X and Y are p-adic Banach spaces over Ωp, we define X ≃Y to mean that X and Y are isometrically isomorphic as p-adic Banach spaces over Ωp. A p- adic Banach algebra over Ωp is a p-adic Banach space A over Ωp such that for x,y ∈A, we have kxykp ≤ kxkpkykp. We shall assume that A has a unit. For x ∈ A, the spectrum σx of x and the resolvent set ρ(x) of x have the usual meaning; ρ(x) is open and hence σx is closed (see [8], Theorem 4, page 114). The resolvent of x is defined by R(z; x) = (z − x)−1, z ∈ ρ(x). If X and Y are p-adic Banach spaces over Ωp, then B(X , Y) is the vector space of Ωp-linear continuous maps form X to Y. B(X , Y) is a p-adic Banach space under the usual operator norm, and B(X ) = B(X , X ) is a p-adic Banach algebra under 32 R.L. Baker this operator norm.

Definition 1.14. Let X be a p-adic Banach space over Ωp. An operator A ∈ B(X ), with compact spectrum σA 6= ∅, is called analytic if the resolvent ∗ R(z; A) is Krasner analytic on σA, in the sense that for all x ∈X and h ∈X , the function z 7→ h(R(z; A)x) is in H0(σA).

Lemma 1.15. Let J be any nonempty indexing set, and define Ωp(J) to be the set of all “sequences” c = (cj)j∈J in Ωp such that for every ǫ > 0 only finitely many |cj|p are > ǫ. Define ||c||p = max |cj |p. Then Ωp(J) is a p-adic j Banach space over Ωp. The notation Ωp(J) is used in [7], page 143. However the usual notation for Ωp(J) is c0(J;Ωp), which is used in [12], page 185. Proof. See [12], Corollary 1, page 185.

Lemma 1.16. Let X ≃ Ωp(J), with J 6= ∅. Let ∅= 6 σ ⊆ Ωp be compact. Let F : σ → B(X ) be an analytic operator-valued function, i.e., for all y ∈ ∗ X , h ∈X the Ωp-valued function

Fh,y(x)= h(F (x)y), x ∈ σ belongs to H0(σ). Let a ∈ Ωp and 0 < r ∈ |Ωp|p. Let Γ ∈ Ωp, with |Γ|p = r. Assume that there is no b ∈ σ such that |b−a|p = r. Finally, let f : Da(r) → Ωp be Krasner analytic on Da(r). Define 1 Sn = f(a + ξΓ)F (a + ξΓ), n = 1, 2,.... n n ξX=1 Then the limit lim Sn n→∞ p6|n exists as an operator in B(X ). We define f(x)F (x) dx to be this operator. a,Γ Moreover, for any y ∈X and h ∈X ∗, we haveR

h f(x)F (x) dx (y) = f(x)Fh,y(x) dx. a,ZΓ   a,ZΓ In particular, assume that an analytic operator A ∈ B(X ) has compact spec- −1 trum σA 6= ∅. Then by definition, the resolvent R(x; A) = (x − A) is Krasner analytic on σA and hence the integral

f(x)(x − a)R(x; A) dx a,ZΓ A CERTAIN p-ADICSPECTRALTHEOREM 33

∗ exists. Moreover, for all y ∈ X and all h ∈ X , if Rh,y(x; A) = h(R(x; A)y), then

h f(x)(x − a)R(x; A) dx (y) = f(x)(x − a)Rh,y(x; A) dx. a,ZΓ   a,ZΓ

Proof. See [7], Lemma 3, page 149.

Definition 1.17. Let A be a p-adic Banach algebra over Ωp. Let A ∈ A have spectrum σA 6= ∅. Define F(A)= L(σA). We now state a “classical” p-adic spectral of M. M. Vishik (see [7], page 149).

Spectral 1.18. Theorem I (Vishik). Let X ≃ Ωp(J), where J is a nonempty indexing set. Let A ∈ B(X ) have compact spectrum σA 6= ∅, and assume that A is analytic. Let f ∈ L(σA). Let 0 < r ∈ |Ωp|p be such that f ∈ Br(σA), and assume that

N N − DσA (r)= Dai (r), σA ⊆ Dai (r ), i i [=1 [=1 where a1, . . . , aN ∈ σA, with the Dai (r) disjoint. Let Γ ∈ Ωp, with |Γ|p = r. Set N f(A)= f(x)(x − ai)R(x; A) dx. i=1 Z Xai,Γ Let g ∈ F(A), and let α, β ∈ Ωp. Then: (1) (αf + βg)(A)= αf(A)+ βg(A);

(2) (f · g)(A)= f(A)g(A);

(3) The mapping f 7→ f(A) is continuous on Br(σA). Proof. See [7], Spectral Theorem, page 149. Definition 1.19. Let A be a p-adic Banach algebra and let X be a p-adic Banach space. Let A ∈ A have compact spectrum σA. Then we say that A is X -analytic iff there exists a unital Ωp-monomorphism θ : A→ B(X ) such that θ(A) is a p-adic Banach subalgebra of B(X ), θ−1 : θ(A) → A is bounded and θ(A) is analytic in B(X ). Such a monomorphism θ : A→ B(X ) is said to be an embedding of A into B(X ). Let B ∈A. Then B is A-analytic iff the following two conditions hold: 34 R.L. Baker

(a)For every r > 0, kR(x; B)kp is bounded on the complement DσB (r) of − DσB (r ).

(b)There exists a sequence (Bk) in A and a sequence (Xk) of p-adic Banach

spaces Xk of the form Xk ≃ Ωp(Jk), with Jk 6= ∅, such that σBk 6= ∅ is compact, Bk is Xk-analytic and lim kB − Bkkp = 0. k→∞

Lemma 1.20. Let A and X be as in Definition 1.19, with X ≃ Ωp(J), where J is a nonempty indexing set. Let A ∈A have compact spectrum σA 6= ∅, and assume that A is X -analytic. Let θ : A → B(X ) be an embedding of A into B(X ), with θ(A) analytic. Let 0 < r ∈ |Ωp|p and Γ ∈ Ωp, with |Γ|p = r. Assume that f ∈ Br(σA) and

N − σA ⊆ Dai (r ), (1) i [=1 where a1, . . . , aN ∈ σA and the Dai (r) are disjoint. Then we have

N −1 1 i i i θ [f(θ(A))] = lim f(xξ)(xξ − ai)R(xξ; A), n→∞ n p6|n i n X=1 ξX=1 N = f(x)(x − ai)R(x; A) dx, (2) i=1 Z Xai,Γ

i where for 1 ≤ i ≤ N and ξ ∈ Ωp, we define xξ = ai + ξΓ. Moreover, the limit in (1) is independent of the Γ, r, a1, . . . , aN . Hence we may define f(A) by f(A)= θ−1[f(θ(A))], and this definition will not depend on θ.

Proof. We have σθ(A) ⊆ σA, hence (1) implies that

N − σθ(A) ⊆ Dai (r ). (3) i [=1

By Lemma 1.3, (3) implies that there exist b1, . . . , bM in σθ(A) and ∅= 6 I ⊆

{ 1,...,N } such that the Dbi (r) are disjoint and

M − − − Dσθ(A) (r )= Dbi (r )= Dai (r ); (4) i i I [=1 [∈ A CERTAIN p-ADICSPECTRALTHEOREM 35

M

Dσθ(A) (r)= Dbi (r)= Dai (r). i i I [=1 [∈ ∗ Let y ∈X and h ∈X be arbitrary. Then by assumption, the function ϕh,y(x)= h(R(x; θ(A))y) is in H0(σθ(A)). Because each ai ∈ σA and f ∈ Br(σA), we see that f is Krasner analytic on each Dai (r). Hence, by Lemma 1.12, it follows from (3) and (4) that f ∈ Br(σθ(A)) and

M N f(x)(x − bi)ϕh,y(x) dx = f(x)(x − ai)ϕh,y(x) dx. (5) i=1 Z i=1 Z X bi,Γ X ai,Γ Hence, by Lemma 1.16, (5) implies that

M h f(x)(x − bi)R(x; θ(A)) dx  i=1 Z X bi,Γ1 N − f(x)(x − ai)R(x; θ(A)) dx (y) = 0. i=1 Z   X ai,Γ2 Because h ∈X ∗ is arbitrary, we see that

M f(x)(x − bi)R(x; θ(A)) dx  i=1 Z X bi,Γ N − f(x)(x − ai)R(x; θ(A)) dx (y) = 0. (6) i=1 Z  X ai,Γ Here we have used the fact that if z ∈ X is such that k(z) = 0 for all k ∈ X ∗, then z = 0 (see [7], Lemma 2, page 144). Finally, because y ∈ X is arbitrary, (6) implies that

M N f(x)(x − bi)R(x; θ(A)) dx = f(x)(x − ai)R(x; θ(A)) dx. (7) i=1 Z i=1 Z X bi,Γ X ai,Γ By (4) and Theorem 1.19, we have

M f(θ(A)) = f(x)(x − bi)R(x; θ(A)) dx, i=1 Z X bi,Γ 36 R.L. Baker consequently, (7) gives

N f(θ(A)) = f(x)(x − ai)R(x; θ(A)) dx. (8) i=1 Z X ai,Γ i Now let 1 ≤ i ≤ N and ξ ∈ Ωp, with |ξ|p = 1. Then xξ = ai + ξΓ ∈ σA, and i therefore R(xξ; A) ∈A, form which it follows that

i i R(xξ; θ(A)) = θ(R(xξ; A)) ∈ θ(A). (9)

Because θ(A) is a p-adic Banach subalgebra of B(X ) and

N 1 i i i f(θ(A)) = lim f(xξ)(xξ − ai)R(xξ; θ(A)), n→∞ n p6|n i n X=1 ξX=1 (9) implies that f(θ(A)) ∈ θ(A). Hence we may define f(A) = θ−1[f(θ(A))]. Because θ−1 is bounded, we have

N −1 −1 θ [f(θ(A))] = θ f(x)(x − ai)R(x; θ(A)) dx  i=1 Z  Xai,Γ N −1 1 i i i = θ lim f(xξ)(xξ − ai)R(xξ; θ(A)) n→∞ n p6|n i ξn  X=1 X=1  N 1 i i −1 i = lim f(xξ)(xξ − ai)θ [R(xξ; θ(A))] n→∞ n p6|n i ξn X=1 X=1 N 1 i i i = lim f(xξ)(xξ − ai)R(xξ; A). n→∞ n p6|n i ξn X=1 X=1 Finally, by Theorem 1.19, the representation

N f(θ(A)) = f(x)(x − ai)R(x; θ(A)) dx i=1 Z Xai,Γ does not depend on Γ, r, a1, . . . , aN . Let A be a p-adic Banach algebra. The following version of Theorem 1.18 applies directly to elements A ∈A that are X -analytic. A CERTAIN p-ADICSPECTRALTHEOREM 37

Spectral 1.21. Theorem II (Vishik). Let X ≃ Ωp(J), where J is a nonempty indexing set. Let A be a p-adic Banach algebra, and let A ∈A have compact spectrum σA 6= ∅, and assume that A is X -analytic. Let f ∈ L(σA). Let 0 < r ∈ |Ωp|p be such that f ∈ Br(σA), and assume that

N N − DσA (r)= Dai (r), σA ⊆ Dai (r ), i i [=1 [=1 where a1, . . . , aN ∈ σA, with the Dai (r) disjoint. Let Γ ∈ Ωp, with |Γ|p = r. Set N f(A)= f(x)(x − ai)R(x; A) dx. i=1 Z Xai,Γ

Let g ∈ F(A), and let α, β ∈ Ωp. Then:

(1) (αf + βg)(A)= αf(A)+ βg(A);

(2) (f · g)(A)= f(A)g(A);

(3) The mapping f 7→ f(A) is continuous on Br(σA).

Proof. Let θ : A → B(X ) be an embedding of A into B(X ), where X ≃ Ωp(J), J 6= ∅. Assume that θ(A) is X -analytic. Then Theorem 1.18 applies to θ(A), and hence Theorem 1.21 follows form Lemma 1.20 and the fact that for −1 f ∈ BσA (r), f(A)= θ [f(θ(A))].

2. Extended p-Adic Spectral Theory

Theorem 2.1. (Perturbation Theory) Let A be a p-adic Banach algebra. Let A ∈A have spectrum σA. Then the following statements hold.

−1 −1 (a)If A is invertible and B ∈A with kA−Bkp < kA kp , then B is invertible and

∞ B−1 = A−1 [(A − B)A−1]n; (1) n=0 X−1 2 −1 −1 kA kpkA − Bkp kB − A kp ≤ −1 . (2) 1 − kA − BkpkA kp 38 R.L. Baker

(b)Let σA ⊆ G ⊆ Ωp. Let ǫ> 0 be a positive real number.

Assume that there exists a constant Nǫ > 0 such that

kR(x; A)kp ≤ Nǫ for all x ∈ G.

Then there exists a δ > 0 such that if B ∈A and ||A − B||p < δ, then

σB ⊆ G (3)

and ||R(x; A) − R(x; B)||p <ǫ for all x ∈ G. (4)

(c)Suppose that there exists an r > 0 such that kR(x; A)kp is bounded on

DσA (r). Moreover, suppose that there exists a sequence (Ak) in A such that each σA 6= ∅ and lim kA − Akkp = 0. Then σA 6= ∅. k k→∞ (d)Let A ∈ A be A-analytic. Then for every ǫ > 0 there exists a δ > 0 such that if B ∈A and ||A − B||p < δ, then

− σB ⊆ DσA (ǫ ) (5)

and

||R(x; A) − R(x; B)||p <ǫ for all x ∈ DσA (ǫ). (6)

Furthermore, σA 6= ∅.

Proof. By Theorem 4, page 108 of [8], if X ∈ A is such that the series ∞ Xn converges, then 1 − X is invertible, with n=0 P ∞ (1 − X)−1 = Xn. (∗) n=0 X To prove (a), suppose that A ∈ A is invertible. Let B ∈ A, with kA − Bkp < −1 −1 kA kp . Then

−1 −1 −1 k1 − BA kp = k(A − B)A kp ≤ kA − BkpkA kp < 1,

∞ and hence the series (1 − BA−1)n is convergent. Therefore (∗) implies that n=0 P 1 − (1 − BA−1)= BA−1 A CERTAIN p-ADICSPECTRALTHEOREM 39 is invertible, with ∞ ∞ (BA−1)−1 = (1 − BA−1)n = [(A − B)A−1]n. n=0 n=0 X X But then because BA−1 is invertible, so is B, and we get ∞ AB−1 = (BA−1)−1 = [(A − B)A−1]n, n X=0 that is, ∞ B−1 = A−1 [(A − B)A−1]n. n=0 X This proves (1). Now, (1) implies that ∞ −1 −1 −1 −1 n −1 kB − A kp = kA [(A − B)A ] − A kp n=0 X∞ −1 −1 n = kA [(A − B)A ] − 1 kp  n=0  ∞X −1 −1 n = kA [(A − B)A ] kp n=1 X ∞ −1 −1 n ≤ kA kpk [(A − B)A ] kp n=1 ∞X −1 −1 n ≤ kA kp k(A − B)kpkA kp n=1 X∞ −1 −1 n = kA kp k(A − B)kpkA kp − 1 n=0  X  1 = kA−1k − 1 p 1 − kA−1k kA − Bk  p p  −1 2 kA kpkA − Bkp = −1 . 1 − kA kpkA − Bkp This proves (2). To prove (b), let ǫ> 0 be arbitrary, and assume that there is a constant Nǫ > 0 such that kR(x; A)kp ≤ Nǫ for all x ∈ G. Define 1 ǫ δ = min , . N N 2 + ǫN  ǫ ǫ ǫ  40 R.L. Baker

Now let B ∈A with kA − Bkp < δ. To prove (3) it suffices to show that G ⊆ ρ(B). (7)

To this end, let x ∈ G. Then kR(x; A)kp ≤ Nǫ, and hence −1 −1 −1 −1 Nǫ ≤ kR(x; A)kp = k(x − A) kp . Therefore,

1 −1 −1 k(x − A) − (x − B)kp = kA − Bkp < ≤ k(x − A) kp . (8) Nǫ It follows from (a) that x − B is invertible, i.e., x ∈ ρ(B). This proves (7). To prove (4), define ǫ µ = 2 . Nǫ + ǫNǫ A short calculation shows that N 2µ ǫ = ǫ. 1 − Nǫµ

Now let x ∈ G. By assumption, kA − Bkp <µ. Because (8) holds, (a) implies that

−1 −1 kR(x; B) − R(x; A)kp = k(x − B) − (x − A) kp −1 2 k(x − A) kpk(x − A) − (x − B)kp ≤ −1 1 − k(x − A) kpk(x − A) − (x − B)kp 2 kR(x; A)k kB − Akp = p 1 − kR(x; A)kpkB − Akp N 2kA − Bk ≤ ǫ p 1 − NǫkA − Bkp N 2µ < ǫ = ǫ. 1 − Nǫµ This proves (4), and hence the proof of (b) is complete. To prove (c), let r> 0 and suppose that kR(x; A)kp is bounded on DσA (r) and suppose that there exists a sequence (Ak) in A such that each σA 6= ∅ and lim kA − Akkp = 0. k k→∞ − − Assume that σA = ∅. Then DσA (r ) = D∅(r ) = ∅ (see Definition 1.1), consequently, DσA (r)=Ωp, from which it follows that kR(x; A)kp is bounded on Ωp. Therefore there exists an Nr > 0 such that

kR(x; A)kp ≤ Nr for all x ∈ Ωp. (9) A CERTAIN p-ADICSPECTRALTHEOREM 41

−1 Now let k be so large that kA − Akkp < Nr . Let x ∈ Ωp = ρ(A) be arbitrary. Then (9) implies that

−1 −1 −1 −1 k(x − Ak) − (x − A)kp = kA − Akkp < Nr ≤ kR(x; A)kp = k(x − A) kp .

It then follows from (a) that x−Ak is invertible, i.e., x ∈ ρ(Ak). Because x ∈ Ωp is arbitrary, we see that ρ(Ak)=Ωp, i.e., σAk = ∅. But this contradicts the assumption that σAk 6= ∅. We conclude that σA 6= ∅. This proves (c). Finally, to prove (d), let A ∈ A be A-analytic. Then by Definition 1.19 the following two conditions hold.

(e)For every r> 0, kR(x; A)kp is bounded on DσA (r).

(f)There exists a sequence (Ak) in A and a sequence (Xk) of p-adic Banach

spaces Xk of the form Xk ≃ Ωp(Jk), with Jk 6= ∅, such that σAk 6= ∅ is compact, Ak is Xk-analytic and lim kA − Akkp = 0. k→∞ − Now let ǫ> 0 be arbitrary. In (b), set G = DσA (r ), then (e) implies that the A and ǫ satisfy the hypothesis of (b), form which we conclude that there exists a δ > 0 such that if B ∈ A and ||A − B||p < δ, then (5) and (6) above hold. Finally, (e) and (f) together imply that A satisfies the hypothesis of (c), and hence σA 6= ∅. This completes the proof of the theorem.

Lemma 2.2. Let A be a p-adic Banach algebra, and let A ∈ A have compact spectrum σA 6= ∅. Let X be a p-adic Banach space such that X ≃ Ωp(J), with J 6= ∅, and suppose that A be X -analytic. Let 0 < r1 ≤ r2, with r1, r2 ∈ |Ωp|p. Let Γ1, Γ2 be in Ωp, with |Γ1|p = r1, |Γ2|p = r2. Assume that

M M − DσA (r1)= Dai (r1), σA ⊆ Dai (r1 ); (1) i i [=1 [=1 N N − DσA (r2)= Dbi (r2), σA ⊆ Dbi (r2 ), (1) i i [=1 [=1 where the Dai (r1), Dbi (r2) are disjoint. If f ∈ Br2 (σA), then f ∈ Br1 (σA), and the following sums exist and are equal.

M N f(x)(x − ai)R(x; A) dx = f(x)(x − bi)R(x; A) dx. (2) i=1 Z i=1 Z X ai,Γ1 X bi,Γ2 42 R.L. Baker

Proof. Let θ : A→ B(X ) be and embedding of A into B(X ) such that θ(A) is analytic. We have σθ(A) ⊆ σA, hence (1) implies that

M N − − σθ(A) ⊆ Dai (r1 ) and σθ(A) ⊆ Dbi (r2 ). (3) i i [=1 [=1

By Lemma 1.3, (3) implies that there exist c1,...,cK ∈ σθ(A) and ∅= 6 I ⊆

{ 1,...,M } such that the Dci (r) are disjoint and

K − − − Dσθ(A) (r1 )= Dci (r1 )= Dai (r1 ) ; (4) i i I [=1 [∈ K

Dσθ(A) (r1)= Dci (r1)= Dai (r1). i i I [=1 [∈

Likewise, by Lemma 1.3, (3) implies that there exist d1,...,dL ∈ σθ(A) and

∅= 6 J ⊆ { 1,...,N } such that the Ddi (r) are disjoint and

L − − − Dσθ(A) (r2 )= Ddi (r2 )= Dbi (r2 ) i i J [=1 [∈ L

Dσθ(A) (r2)= Ddi (r2)= Dbi (r2). (5) i i J [=1 [∈

Now let f ∈ Br2 (σA). Let a ∈ Ωp, with Da(r) ⊆ Dσθ(A) (r2). Then (1) and (5) imply that

Da(r) ⊆ Dσθ(A) (r2)= Dbi (r2) ⊆ DσA (r2). i J [∈

It follows that f is Krasner analytic on Da(r), and consequently, f ∈ Br2 (σθ(A)).

Because Br2 (σθ(A)) ⊆ Br1 (σθ(A)), we see that f ∈ Br1 (σθ(A)). Let y ∈ X and ∗ h ∈ X . Then the function ϕh,y(x) = h(R(x; θ(A))y) is in H0(σθ(A)). By Lemma 1.11, (4) and (5) imply that

K L f(x)(x − ci)ϕh,y(x) dx = f(x)(x − di)ϕh,y(x) dx. (6) i=1 Z i=1 Z X ci,Γ1 X di,Γ2 Hence, by Lemma 1.16, (6) implies that A CERTAIN p-ADICSPECTRALTHEOREM 43

K h f(x)(x − ci)R(x; θ(A)) dx  i=1 Z X ci,Γ1 L − f(x)(x − di)R(x; θ(A)) dx (y) = 0. i=1 Z   X di,Γ2 Because h ∈X ∗ is arbitrary, we see that

K f(x)(x − ci)R(x; θ(A)) dx  i=1 Z X ci,Γ L − f(x)(x − di)R(x; θ(A)) dx (y) = 0. (7) i=1 Z  X di,Γ Here, again, we have used the fact that if z ∈ X is such that k(z) = 0 for all k ∈ X ∗, then z = 0 (see [7], Lemma 2, page 144). Finally, because y ∈ X is arbitrary, (7) implies that

K L f(x)(x − ci)R(x; θ(A)) dx = f(x)(x − di)R(x; θ(A)) dx. (8) i=1 Z i=1 Z X ci,Γ X di,Γ Similarly, we see that by Lemma 1.12 and Lemma 1.16, (3), (4), and (5) together imply that

K M f(x)(x − ci)R(x; θ(A)) dx = f(x)(x − ai)R(x; θ(A)) dx; i=1 Z i=1 Z X ci,Γ1 X ai,Γ1 L N f(x)(x − di)R(x; θ(A)) dx = f(x)(x − bi)R(x; θ(A)) dx. (9) i=1 Z i=1 Z X di,Γ2 X bi,Γ2 Statements (8) and (9) then imply that

M N f(x)(x − ai)R(x; θ(A)) dx = f(x)(x − bi)R(x; θ(A)) dx (10). i=1 Z i=1 Z X ai,Γ1 X bi,Γ2 i Let ξ ∈ Ωp be arbitrary with |ξ|p = 1; for 1 ≤ i ≤ M define xξ = ai + i i i ξΓ1; for 1 ≤ i ≤ N define yξ = bi + ξΓ2. Then xξ,yξ ∈ σA, and therefore 44 R.L. Baker

i i R(xξ; A),R(yξ; A) ∈A, form which it follows that

i i R(xξ; θ(A)) = θ(R(xξ; A)) ∈ θ(A); (11) i i R(yξ; θ(A)) = θ(R(yξ; A)) ∈ θ(A). We have M M 1 i i i f(x)(x − ai)R(x; θ(A))dx = lim f(xξ)(xξ − ai)R(xξ; θ(A)); n→∞ n i=1 Z p6|n i=1 ξn=1 Xai,Γ1 X X (12) N N 1 i i i f(x)(x − bi)R(x; θ(A))dx = lim f(yξ)(yξ − bi)R(yξ; θ(A)). n→∞ n i=1 Z p6|n i=1 ξn=1 Xbi,Γ2 X X Because θ(A) is a p-adic Banach subalgebra of B(X ) and θ−1 : θ(A) → A is bounded, (11) and (12) imply that the following sums exist and are equal.

M N f(x)(x − ai)R(x; A) dx = f(x)(x − bi)R(x; A) dx. i=1 Z i=1 Z X ai,Γ1 X bi,Γ2 This proves statement (2), and hence the proof of the lemma is complete. Lemma 2.3. Let A be a p-adic Banach algebra, and let A ∈ A have compact spectrum σA 6= ∅. Let X be a p-adic Banach space such that X ≃ Ωp(J), with J 6= ∅, and suppose that A be X -analytic. Let 0 < r ∈ |Ωp|p, and let Γ ∈ Ωp, with |Γ|p = r. Assume that

M M − DσA (r)= Dbi (r), σA ⊆ Dbi (r ), (1) i i [=1 [=1 where b1, . . . , bM are in Ωp and the Dbi (r) are disjoint. Finally, suppose that

N − σA ⊆ Dai (r ), (2) i [=1 where a1, . . . , aN are in Ωp and the Dai (r) are disjoint. If f is Krasner analytic on each Dai (r), then we have f ∈ Br(σA) and

M N f(x)(x − bi)R(x; A) dx = f(x)(x − ai)R(x; A) dx. i=1 Z i=1 Z X bi,Γ X ai,Γ A CERTAIN p-ADICSPECTRALTHEOREM 45

Proof. Let θ : A→ B(X ) be and embedding of A into B(X ) such that θ(A) is analytic. We have σθ(A) ⊆ σA, hence (1) and (2) imply that

M N − − σθ(A) ⊆ Dbi (r ) and σθ(A) ⊆ Dai (r ). (3) i i [=1 [=1 By Lemma 1.3, (3) implies that there exist c1,...,cL in σθ(A) and ∅= 6 I ⊆

{ 1,...,N } such that the Dci (r) are disjoint and

L − − − Dσθ(A) (r )= Dci (r )= Dai (r ); (4) i i I [=1 [∈ L

Dσθ(A) (r)= Dci (r)= Dai (r). i i I [=1 [∈

Now let f be Krasner analytic on each Dai (r). Let a ∈ Ωp, with Da(r) ⊆

DσA (r). Because a ∈ DσA (r) and σA is compact, there exists b ∈ σA such that |a − b|p ≤ r. Then by (2) there exists 1 ≤ i ≤ N such that |b − ai|p < r. We then have |a − ai|p ≤ max{ |a − b|p, |b − ai|p }≤ r.

Therefore a ∈ Dai (r), and hence Da(r)= Dai (r). Because f is Krasner analytic on Dai (r), Lemma 1.4 implies that f is Krasner analytic on Da(r). This proves ∗ that f is in Br(σA). Let y ∈X and h ∈X be arbitrary. Then by assumption, the function ϕh,y(x) = h(R(x; θ(A))y) is in H0(σθ(A)). Now, by (1), for all

1 ≤ i ≤ M, we have Dbi (r) ⊆ DσA (r), and hence f is Krasner analytic on

Dbi (r). Because each ci ∈ σθ(A), it follows from (3), (4) and Lemma 1.12 that

L M f(x)(x − ci)ϕh,y(x) dx = f(x)(x − bi)ϕh,y(x) dx; (5) i=1 Z i=1 Z X ci,Γ X bi,Γ L N f(x)(x − ci)ϕh,y(x) dx = f(x)(x − ai)ϕh,y(x) dx. i=1 Z i=1 Z X ci,Γ X ai,Γ We conclude from (5) that

M N f(x)(x − bi)ϕh,y(x) dx = f(x)(x − ai)ϕh,y(x) dx. (6) i=1 Z i=1 Z X bi,Γ X ai,Γ 46 R.L. Baker

Hence, by Lemma 1.16, (6) implies that

M h f(x)(x − bi)R(x; θ(A)) dx  i=1 Z X bi,Γ1 N − f(x)(x − ai)R(x; θ(A)) dx (y) = 0. i=1 Z   X ai,Γ2 Because h ∈X ∗ is arbitrary, we see that

M f(x)(x − bi)R(x; θ(A)) dx  i=1 Z X bi,Γ N − f(x)(x − ai)R(x; θ(A)) dx (y) = 0. (7) i=1 Z  X ai,Γ Here, once more, we have used the fact that if z ∈X is such that k(z)=0 for all k ∈X ∗, then z = 0 (see [7], Lemma 2, page 144). Finally, because y ∈X is arbitrary, (7) implies that M N f(x)(x − bi)R(x; θ(A)) dx = f(x)(x − ai)R(x; θ(A)) dx. (8) i=1 Z i=1 Z X bi,Γ X ai,Γ Now, for x ∈ ρ(A), we have x ∈ ρ(θ(A)), with θ((x−A)−1) = (x−θ(A))−1, and hence (x − θ(A))−1 ∈ θ(A). Because θ−1 : θ(A) → A is bounded, (8) implies that the following sums exist and are equal. M N f(x)(x − bi)R(x; A) dx = f(x)(x − ai)R(x; A) dx. i=1 Z i=1 Z X bi,Γ X ai,Γ This completes the proof of the lemma. Lemma 2.4. Let A be a p-adic Banach algebra. Let A ∈A be A-analytic. Let (Ak) be a sequence in A and (Xk) a sequence of p-adic Banach spaces Xk of the form Xk ≃ Ωp(Jk), with Jk 6= ∅, such that σAk 6= ∅ is compact, Ak is Xk-analytic and lim kA−Akkp = 0 (see Definition 1.19). Let A have spectrum k→∞ σA. Let 0 < r ∈ |Ωp|p, and let Γ ∈ Ωp, with |Γ|p = r. Assume that N − σA ⊆ Dai (r ), (1) i [=1 A CERTAIN p-ADICSPECTRALTHEOREM 47

where a1, . . . , aN ∈ Ωp and the Dai (r) are disjoint. Assume that for each k there exists ak,1, . . . , ak,Nk in σAk such that

Nk D (r)= D (r); σAk ak,i i [=1 Nk − σAk ⊆ Dak,i (r ), i [=1 where the Dak,i (r) are disjoint. If f is Krasner analytic on the Dai (r),then the following limits exist and are equal.

N Nk lim f(x)(x−ai)R(x; Ak) dx = lim f(x)(x−ak,i)R(x; Ak) dx. k→∞ k→∞ i=1 Z i=1 Z X ai,Γ X ak,i,Γ

Proof. Because A is A-analytic, (d) of Theorem 2.1 implies that there exists − a δ0 > 0 such that if B ∈ A and kA − Bkp < δ0, then σB ⊆ DσA (r ). By (1) we have N − − DσA (r ) ⊆ Dai (r ). (3) i [=1 Let k0 be so large that for k ≥ k0, we have kA − Akkp < δ0. Then for k ≥ k0, (3) implies that N − σAk ⊆ Dai (r ). (4) i [=1 For the rest of the proof we may assume that for all k, k ≥ k0. Because f is

Krasner analytic on each Dai (r), it follows from (2), (4), and Lemma 2.3 that for all k, f is in Br(σAk ) and

N Nk f(x)(x − ai)R(x; Ak) dx = f(x)(x − ak,i)R(x; Ak) dx. (5) i=1 Z i=1 Z X ai,Γ X ak,i,Γ

Also, because f is Krasner analytic on each Dai (r), Lemma 1.5 implies that for each 1 ≤ i ≤ N, max |f(x)|p is attained when |x − ai|p = r, and hence we x∈Dai (r) may write ρi = max |f(x)|p = max |f(x)|p. x∈Dai (r) |x−ai|p=r 48 R.L. Baker

Now let 0 < ǫ < r be arbitrary. Let µ> 0 be a positive number such that 1 µ + max ρj > . j r

By (d) of Theorem 2.1 there exists a positive number δ1 > 0 such that if B ∈A and kA − Bkp < δ1, then − σB ⊆ DσA (ǫ ) (6) and ǫ kR(x; A) − R(x; B)kp < for all x ∈ DσA (ǫ). r(µ + max ρj) j

Now let k1 ≥ k0 be so large that kA − Akkp < δ1 for k ≥ k1. Then for k ≥ k1, − − we have σAk ⊆ DσA (ǫ ) ⊆ DσA (r ). Therefore (5) implies that, for all k, l ≥ k1 i we get, with xξ = ai + ξΓ,

Nk Nl f(x)(x − ak,i)R(x; Ak) dx − f(x)(x − al,i)R(x; Al) dx = i=1 Z i=1 Z Xak,i,Γ Xal,i,Γ (7) N N f(x)(x − ai)R(x; Ak) dx − f(x)(x − ai)R(x; Al) dx = i=1 Z i=1 Z Xai,Γ Xai,Γ N 1 i i i i lim f(xξ)(xξ − ai)(R(xξ; Ak) − R(xξ; A))− n→∞ n p6|n i n  X=1 ξX=1 N 1 f(xi )(xi − a )(R(xi ; A ) − R(xi ; A)) . n ξ ξ i ξ l ξ i n X=1 ξX=1  Now, we claim that for 1 ≤ i ≤ N and ξ ∈ Ωp, with, |ξ|p = 1, i xξ ∈ DσA (ǫ). (8) i − i − To prove this claim, suppose that xξ ∈ DσA (ǫ ), then xξ ∈ DσA (r ), hence (3) i − implies that for some 1 ≤ j ≤ N, xξ ∈ Daj (r ). We must have i = j, for if i 6= j, i i then because |xξ − ai|p = |ξΓ|p = r, we would have xξ ∈ Dai (r) ∩ Daj (r) = i − ∅. Thus, i = j, which implies that xξ ∈ Dai (r ). But then we have the i contradiction that r = |xξ − ai|p < r. Thus (8) holds. If p 6 | n, 1 ≤ i ≤ N and m ≥ k0, then (6) and (8) imply that 1 f(xi )(xi − a )(R(xi ; A ) − R(xi ; A)) (9) n ξ ξ i ξ m ξ ξn=1 p X

A CERTAIN p-ADICSPECTRALTHEOREM 49

i i ≤ max ρj · |Γ|p · max kR(xξ; Am) − R(xξ; A)kp j ξn=1 ǫ ≤(r max ρj) j r(µ + max ρj)  j  <ǫ.

Therefore, (7) and (9) imply that for k, l ≥ k1,

Nk Nl f(x)(x − ak,i)R(x; Ak) dx − f(x)(x − al,i)R(x; Al) dx ≤ ǫ. i=1 Z i=1 Z p Xak,i,Γ Xal,i,Γ

Because 0 < ǫ < r is arbitrary, we see that the terms

Nk N f(x)(x − ak,i)R(x; Ak) dx = f(x)(x − ai)R(x; Ak) dx i=1 Z i=1 Z Xak,i,Γ Xai,Γ form a Cauchy sequence. Hence the following limits exist and are equal.

Nk N lim f(x)(x − ak,i)R(x; Ak) dx = lim f(x)(x − ai)R(x; Ak) dx. k→∞ k→∞ i=1 Z i=1 Z Xak,i,Γ Xai,Γ This completes the proof of the lemma. Lemma 2.5. Let A be a p-adic Banach algebra. Let A ∈A be A-analytic. Let (Ak) be a sequence in A and (Xk) a sequence of p-adic Banach spaces Xk of the form Xk ≃ Ωp(Jk), with Jk 6= ∅, such that σAk 6= ∅ is compact, Ak is Xk-analytic and lim kA − Akkp = 0. Let A have spectrum σA. Let 0 < r2 ≤ r1 k→∞ be in |Ωp|p, and let Γ1, Γ2 be in Ωp, with |Γ1|p = r1, |Γ2|p = r2. Assume that a1, . . . , aM ; b1, . . . , bN ∈ Ωp are given, with

M N − − σA ⊆ Dai (r1 ) and σA ⊆ Dbi (r2 ), (1) i i [=1 [=1 where the Dai (r1) are disjoint and the Dbi (r2) are disjoint. Let f be Krasner analytic on the Dai (r1) and on the Dbi (r2). Then the following limits exist and are equal:

M N lim f(x)(x − ai)R(x; Ak) dx = lim f(x)(x − bi)R(x; Ak) dx. k→∞ k→∞ i=1 Z i=1 Z X ai,Γ1 X bi,Γ2 50 R.L. Baker

Proof. Because A is A-analytic, (d) of Theorem 2.1 implies that σA 6= ∅ and that there exists a δ > 0 such that if B ∈A with kA − Bkp < δ, then − σB ⊆ DσA (ri ), for i = 1, 2.

Hence, because lim kA − Akkp = 0, by (1) we may assume that for all k, k→∞ M N − − σAk ⊆ Dai (r1 ) and σAk ⊆ Dbi (r2 ). (2) i i [=1 [=1 Then by Lemma 1.3, (2) implies that for all k there exist Ik, Jk 6= ∅ such that Ik ⊆ { 1,...,M }, Jk ⊆ { 1,...,N } and σ ⊆ D (r−)= D (r−) and D (r )= D (r ); (3) Ak σAk 1 ai 1 σAk 1 ai 1 i I i I [∈ k [∈ k σ ⊆ D (r−)= D (r−) and D (r )= D (r ). Ak σAk 2 bi 2 σAk 2 bi 2 i J i J [∈ k [∈ k Now let k be arbitrary. By Lemma 2.3, (2) and (3) together imply that M f(x)(x − ai)R(x; Ak) dx = f(x)(x − ai)R(x; Ak) dx; (4) i=1 Z i∈Ik Z X ai,Γ1 X ai,Γ1 N f(x)(x − bi)R(x; Ak) dx = f(x)(x − bi)R(x; Ak) dx. i=1 Z i∈Jk Z X bi,Γ2 X bi,Γ2 By Lemma 2.2, (2) and (3) together also imply that

f(x)(x − ai)R(x; Ak) dx = f(x)(x − bi)R(x; Ak) dx. (5) i∈Ik Z i∈Jk Z X ai,Γ1 X bi,Γ2 Consequently, (4) and (5) together imply that for all k, M N f(x)(x − ai)R(x; Ak) dx = f(x)(x − bi)R(x; Ak) dx. (6) i=1 Z i=1 Z X ai,Γ1 X bi,Γ2 It follows from (1), (3), (6), and Lemma 2.4 that the following limits exist and are equal. M N lim f(x)(x − ai)R(x; Ak) dx = lim f(x)(x − bi)R(x; Ak) dx. k→∞ k→∞ i=1 Z i=1 Z X ai,Γ1 X bi,Γ2 A CERTAIN p-ADICSPECTRALTHEOREM 51

This completes the proof of the lemma. Lemma 2.6. Let A be a p-adic Banach algebra. Let A ∈A be A-analytic with spectrum σA. Let (Ak) be a sequence in A and (Xk) a sequence of p-adic

Banach spaces Xk of the form Xk ≃ Ωp(Jk), with Jk 6= ∅, such that σAk 6= ∅ is compact, Ak is Xk-analytic and lim kA − Akkp = 0. Let 0 < r ∈ |Ωp|p, and let k→∞ Γ ∈ Ωp, with |Γ|p = r. Assume that a1, . . . , aN ∈ Ωp are given, with

N − σA ⊆ Dai (r ), (1) i [=1 where the Dai (r) are disjoint. Assume that f is Krasner analytic on the Dai (r). Then the following limits exist and are equal.

N N 1 i i i lim f(xξ)(xξ − ai)R(xξ; A) = lim f(x)(x − ai)R(x; Ak) dx, n→∞ n k→∞ p6|n i=1 ξn=1 i=1 Z X X X ai,Γ

i where for each 1 ≤ i ≤ N and ξ ∈ Ωp, we set xξ = ai + ξΓ. Therefore the sum

N f(x)(x − ai)R(x; A) dx i=1 Z X ai,Γ exits and we have

N N f(x)(x − ai)R(x; A) dx = lim f(x)(x − ai)R(x; Ak) dx. k→∞ i=1 Z i=1 Z X ai,Γ X ai,Γ

Proof. Because f is Krasner analytic on each of the Dai (r), it follows from Lemma 1.5 that for 1 ≤ i ≤ N, max |f(x)|p is attained when |x − ai|p = r, x∈Dai (r) and hence we may write

ρi = max |f(x)|p = max |f(x)|p. x∈Dai (r) |x−ai|p=r

Now let 0 < ǫ < r be arbitrary. Let µ> 0 be a positive number such that 1 µ + max ρj > . j r 52 R.L. Baker

Because A is A-analytic, (d) of Theorem 2.1 implies that σA 6= ∅ and that there exists a positive number δ > 0 such that if B ∈A and kA − Bkp < δ, then

− σB ⊆ DσA (ǫ ) and ǫ kR(x; A) − R(x; B)kp < for all ∈ DσA (ǫ). (2) r(µ + max ρj) j

Then by Lemma 1.3, (1) implies that for all k there exist Ik 6= ∅ such that Ik ⊆ { 1,...,N } and

σ ⊆ D (r−)= D (r−) and D (r)= D (r). (3) Ak σAk ai σAk ai i I i I [∈ k [∈ k By Lemma 2.4, (1) and (3) together imply that the following limits exist and are equal.

N N lim f(x)(x − ai)R(x; Ak) dx = lim f(x)(x − ai)R(x; Ak) dx. k→∞ k→∞ i=1 Z i∈Ik Z X ai,Γ X ai,Γ

Now let k0 be so large that kA − Ak0 kp < δ and

N N f(x)(x − ai)R(x; Ak ) dx − lim f(x)(x − ai)R(x; Ak) dx < ǫ. 0 k→∞ i=1 Z i=1 Z p Xai,Γ Xai,Γ

(4) Observe that if 1 ≤ i ≤ N and ξ ∈ Ωp, with |ξ|p = 1, then as in statement (8) i of Lemma 2.4, we have xξ = ai + ξΓ ∈ DσA (ǫ). Now let n0 be so large that if n ≥ n0 and p 6 | n, then

N 1 f(xi )(xi − a )R(xi ; A ) − f(x)(x − a )R(x; A ) dx < ǫ. n ξ ξ i ξ k0 i k0 i=1  ξn=1 Z  p X X ai,Γ

(5) For n ≥ n0 and p 6 | n, we have

N N 1 i i i f(xξ)(xξ − ai)R(xξ; A) − lim f(x)(x − ai)R(x; Ak) dx (6) n k→∞ i=1 ξn=1 i=1 Z X X Xai,Γ N 1 = f(xi )(xi − a ) R(xi ; A) − R(xi ; A ) + n ξ ξ i ξ ξ k0 i=1 ξn=1 X X  A CERTAIN p-ADICSPECTRALTHEOREM 53

N 1 f(xi )(xi − a )R(xi ; A ) − f(x)(x − a )R(x; A ) dx + n ξ ξ i ξ k0 i k0 i=1  ξn=1 Z  X X ai,Γ N N f(x)(x − ai)R(x; Ak ) dx − lim f(x)(x − ai)R(x; Ak) dx. 0 k→∞ i=1 Z i=1 Z Xai,Γ Xai,Γ

Thus, for n ≥ n0 and p 6 | n, (1)-(6) together imply that

N N 1 i i i f(xξ)(xξ −ai)R(xξ; A)− lim f(x)(x−ai)R(x; Ak) dx < ǫ. n k→∞ i=1 ξn=1 i=1 Z p X X Xai,Γ

Because ǫ> 0 is arbitrary, this completes the proof of the lemma. Definition 2.7. Let A be a p-adic Banach algebra. Let A ∈ A be A- analytic with spectrum σA. Let (Ak) be a sequence in A and (Xk) a sequence of p-adic Banach spaces Xk of the form Xk ≃ Ωp(Jk), with Jk 6= ∅, such that

σAk 6= ∅ is compact, Ak is Xk-analytic and lim kA − Akkp = 0. Let f ∈ F(A). k→∞ Let 0 < r ∈ |Ωp|p, and let Γ ∈ Ωp, with |Γ|p = r. Assume that a1, . . . , aN ∈ Ωp are given, with N − σA ⊆ Dai (r ), (1) i [=1 where the Dai (r) are disjoint. Assume that f is Krasner analytic on the Dai (r). By Lemma 2.6, (1) implies that

N N f(x)(x − ai)R(x; A) dx = lim f(x)(x − ai)R(x; Ak) dx. k→∞ i=1 Z i=1 Z X ai,Γ X ai,Γ Now, Lemma 2.5 implies that

N lim f(x)(x − ai)R(x; Ak) dx k→∞ i=1 Z X ai,Γ does not depend on a1, . . . , aN , r, Γ, hence we may define f(A) by

N f(A)= f(x)(x − ai)R(x; A) dx. i=1 Z X ai,Γ 54 R.L. Baker

Lemma 2.8. Let A be a p-adic Banach algebra. Let A ∈A have spectrum σA, and assume that A is A-analytic. Let f ∈ F(A). Let 0 < r ∈ |Ωp|p, and let Γ be in Ωp, with |Γ|p = r. Assume that a1, . . . , aN in Ωp are given, with

N − σA ⊆ Dai (r ), (1) i [=1 where the Dai (r) are disjoint and f is Krasner analytic on the Dai (r). Then for any ǫ > 0 there exists a δ > 0 such that if B ∈ A is A-analytic with kA − Bkp < δ, then N

σB ⊆ Dai (r), i [=1 f ∈ F(B) and kf(A) − f(B)kp <ǫ.

Proof. Because f is Krasner analytic on each Dai (r), Lemma 1.5 implies that for each 1 ≤ i ≤ N, max |f(x)|p is attained when |x − ai|p = r, and x∈Dai (r) hence we may write

ρi = max |f(x)|p = max |f(x)|p. x∈Dai (r) |x−ai|p=r

Now let 0 < ǫ < r be arbitrary. Let µ> 0 be a positive number such that 1 µ + max ρj > . j r

Because A is A-analytic, (d) of Theorem 2.1 implies that σA 6= ∅ and that there exists a positive number δ > 0 such that if B ∈A and kA − Bkp < δ, then

− σB ⊆ DσA (ǫ ) (2) and ǫ kR(x; A) − R(x; B)kp < for all x ∈ DσA (ǫ). (3) r(µ + max ρj) j

Now assume that B ∈ A is A-analytic, and that kA − Bkp < δ. It is easy to see that (1) implies N − − DσA (r ) ⊆ Dai (r ). (4) i [=1 A CERTAIN p-ADICSPECTRALTHEOREM 55

− Then by (2), we have σB ⊆ DσA (ǫ ), and hence (4) implies that

N − − − σB ⊆ DσA (ǫ ) ⊆ DσA (r ) ⊆ Dai (r ). (5) i [=1

Because B is A-analytic, σB 6= ∅, and hence (5) implies that f ∈ F(B). Then by (1), (5) and Definition 2.7, f(A) and f(B) are well defined, with

N N 1 i i i f(A)= f(x)(x − ai)R(x; A) dx = lim f(xξ)(xξ − ai)R(xξ; A); n→∞ n i=1 Z p6|n i=1 ξn=1 Xai,Γ X X N N 1 i i i f(B)= f(x)(x − ai)R(x; B) dx = lim f(xξ)(xξ − ai)R(xξ; B). n→∞ n i=1 Z p6|n i=1 ξn=1 Xai,Γ X X Then we get

N 1 i i i i f(A) − f(B) = lim f(xξ)(xξ − ai) R(xξ; A) − R(xξ; B) . (6) n→∞ n p6|n i=1 ξn=1 X X  

Now, we claim that for 1 ≤ i ≤ N and ξ ∈ Ωp, with, |ξ|p = 1,

i xξ ∈ DσA (ǫ). (7)

The proof of this claim is similar to the proof of statement (8) in Lemma 2.4. It then follows from (3), (6) and (7) that

N 1 i i i i kf(A) − f(B)kp = lim f(xξ)(xξ − ai) R(xξ; A) − R(xξ; B) n→∞ n p6|n i=1 ξn=1 p X X   ǫ ≤ (r max ρj) j r(µ + max ρj)  j  < ǫ.

This completes the proof of the lemma. Lemma 2.9. Let A be a p-adic Banach algebra. Let A ∈A be A-analytic with spectrum σA. Let (Ak) be a sequence in A and (Xk) a sequence of p-adic

Banach spaces Xk of the form Xk ≃ Ωp(Jk), with Jk 6= ∅, such that σAk 6= ∅ is 56 R.L. Baker

compact, Ak is Xk-analytic and lim kA − Akkp = 0. Let 0 < r ∈ |Ωp|p, and let k→∞ Γ ∈ Ωp, with |Γ|p = r. Assume that a1, . . . , aN ∈ Ωp are given, with

N − σA ⊆ Dai (r ), (1) i [=1 where the Dai (r) are disjoint. Assume that f is Krasner analytic on the Dai (r).

Then there exists a k0 such that for all k ≥ k0, f ∈ Br(σAk ) and

f(A) = lim f(Ak), k→∞ where f(Ak) is given by Theorem 1.21. Proof. Because A is A-analytic, (d) of Theorem 2.1 implies that there exists − a δ > 0 such that if B ∈A and kA − Bkp < δ, then σB ⊆ DσA (r ). By (1) we have N − − DσA (r ) ⊆ Dai (r ). (2) i [=1 Let k0 be so large that for k ≥ k0, we have kA − Akkp < δ. Then for k ≥ k0, (2) implies that N − σAk ⊆ Dai (r ). (3) i [=1 For the rest of the proof we may assume that for all k, k ≥ k0. Then by Lemma

1.3, (3) implies that for all k there exists ak,1, . . . , ak,Nk in σAk such that the D (r),...,D (r) are disjoint with ak,1 ak,Nk

Nk Nk D (r−)= D (r−) and D (r)= D (r). (4) σAk ak,i σAk ak,i i i [=1 [=1 It follows from (3), (4), and Lemma 2.4 that

N f(A) = lim f(x)(x − ai)R(x; Ak) dx (5) k→∞ i=1 Z X ai,Γ Nk = lim f(x)(x − ak,i)R(x; Ak) dx. k→∞ i=1 Z X ak,i,Γ A CERTAIN p-ADICSPECTRALTHEOREM 57

Now let k ≥ k . We claim that f ∈ B (σ ). To see this let D (r) ⊆ D (r). 0 r Ak a σAk Then by (4) there exists 1 ≤ i ≤ Nk such that |a − ak,i|p ≤ r. Then by (3) There exists 1 ≤ j ≤ N such that |ak,i − aj|p < r. It follows that |a − aj|p ≤ max |ak,i − aj|p, |a − aj|p ≤ r, and hence we have a ∈ Daj (r). Because f is

Krasner analytic on Daj (r), Lemma 1.4 implies that f is Krasner analytic on Da(r). This proves the claim. Therefore by (4) and Theorem 1.21 we see that

Nk f(Ak)= f(x)(x − ai)R(x; Ak) dx. i=1 Z X ak,i,Γ

Hence (5) implies that

f(A) = lim f(Ak). k→∞ This completes the proof of the lemma. Extended 2.10. p-adic Spectral Theorem. Let A be a p-adic Banach algebra. Let A ∈ A be A-analytic, with spectrum σA. Let f, g ∈ F(A). Let 0 < r ∈ |Ωp|p, and let Γ be in Ωp, with |Γ|p = r. Assume that a1, . . . , aN in Ωp are given, with N − σA ⊆ Dai (r ), (1) i [=1 where the Dai (r) are disjoint. Finally, suppose that f and g are Krasner analytic on each Dai (r). By (1) and Definition 2.7, f(A) and g(A) are well-defined, with

N f(A)= f(x)(x − ai)R(x; A) dx, i=1 Z Xai,Γ N g(A)= g(x)(x − ai)R(x; A) dx. i=1 Z Xai,Γ

Let α, β ∈ Ωp. Then αf + βg,f · g ∈ F(A), and

(2) (αf + βg)(A)= αf(A)+ βg(A);

(3) (f · g)(A)= f(A)g(A);

(4) The mapping h : Br,a1,...,aN →A given by h 7→ h(A) is norm continuous. 58 R.L. Baker

Proof. Let α, β ∈ Ωp. Then it is clear that αf + βg,f · g ∈ F(A). Because A is A-analytic, there exists a sequence (Ak) in A and a sequence (Xk) of p-adic

Banach spaces Xk of the form Xk ≃ Ωp(Jk), with Jk 6= ∅, such that σAk 6= ∅ is compact, Ak is Xk-analytic and lim kA − Akkp = 0. Then by Lemma 2.9 and k→∞ Theorem 1.21 we have

(αf + βg)(A) = lim (αf + βg)(Ak) k→∞

= lim (αf(Ak)+ βg(Ak)) k→∞

= α lim f(Ak)+ β lim g(Ak) k→∞ k→∞ = αf(A)+ βg(A); and we also have

(f · g)(A) = lim (f · g)(Ak) k→∞

= lim f(Ak)g(Ak) k→∞ = f(A)g(A).

This proves (2) and (3). To prove (4), let h0 ∈ Br,a1,...,aN . Because A is A- analytic, by Definition 1.19 there exists a constant Nr > 0 such that kR(x; A)kp ≤ ǫ Nr for all x ∈ DσA (r). Now let ǫ > 0 be given, and define δ = . Let rNr h ∈ Br,a1,...,aN , with kh − h0ku < δ. For 1 ≤ i ≤ N and ξ ∈ Ωp, with |ξ|p = 1, i i set xξ = ai + ξΓ. Then for 1 ≤ i ≤ N and |ξ|p = 1, xξ ∈ DσA (r). Hence by (2) and Lemma 2.6, we have

kh(A) − h0(A)kp =k(h − h0)(A)kp N 1 i i i i = lim [h(xξ) − h0(xξ)](xξ − ai)R(xξ; A) n→∞ n p6|n i=1 ξn=1 p X X N 1 i i i = lim [h(xξ) − h0(xξ)](ξΓ)R(xξ; A) n→∞ n p6|n i=1 ξn=1 p X X

≤kh − h0ku(rNr) ǫ < (rNr)= ǫ. rNr

This proves (4). A CERTAIN p-ADICSPECTRALTHEOREM 59

Definition 2.11. Let X be a p-adic Banach space over Ωp. Let ∅= 6 X ⊆X . ∗ Then X is weakly bounded iff for each h ∈ X the set { |h(x)|p | x ∈ X } is bounded.

Lemma 2.12. Let X ≃ Ωp(J), where J 6= ∅. Let ∅= 6 X ⊆ X be weakly bounded. Then the set { kxkp | x ∈ X } is bounded.

Proof. For each x = (xj)j∈J ∈ Ωp(J) and each ǫ> 0 let

Ix = { j ∈ J | xj 6= 0 } and Ix(ǫ)= { j ∈ J | kxj kp >ǫ }.

By definition of Ωp(J), for each x ∈ Ωp(J) and ǫ > 0 the set Ix(ǫ) is finite; ∞ 1 and because Ix = Ix( n ), we see that Ix is countable. Now assume that the n=1 { kxkp | x ∈ X } is notS bounded. Then we can find a sequence (xn) in X such that lim kxnkp = ∞. n→∞ ∞ Let I be the n=1 Ixn , and define Y = { y ∈ Ωp(J) |,Iy ⊆ I }. Then Y is a p-adic Banach subspace Ωp(J) such that X ⊆ Y. Moreover, the S mapping θ : Y → Ωp(I) defined by y = (yj)j∈J 7→ (yj)j∈I ∈ Ωp(I) is an isometric isomorphism of Y onto Ωp(I). Because each xn ∈ Y we may define a sequence (yn) in Ωp(I) by yn = θ(xn). Then

lim kynkp = lim kθ(xn)kp = lim kxnkp = ∞. n→∞ n→∞ n→∞

∗ ∗ Let g ∈ Ωp(I) , and define h ∈X by

h(x)= g((xj )j∈I), x = (xj)j∈J ∈X .

Write xn = (xnj)j∈J , then we have

g(yn)= g(θ(xn)) = g((xnj)j∈I )= h((xnj)j∈J )= h(xn).

By assumption the sequence (|h(xn)|p) is bounded, and hence the sequence (|g(yn)|p) is bounded. Because I is countably infinite we may identify I with the set N of positive integers. We have that kynkp = max |ynj|p. Let αn denote j the first j for which |ynj|p = kynkp, and let βn denote the last j for which |ynj|p = kynkp. We have the following two cases to consider. 60 R.L. Baker

Case 1. In this case the sequence (αn) is bounded. Then there exists some ∗ j0 such that αn = j0 for infinitely many n. Let h ∈ Ωp(N) be defined by

h((zj )) = zj0 , (zj ) ∈ Ωp(N).

Then for infinitely many n we have

|h(yn)|p = |ynj0 |p = |ynαn |p = kynkp.

Because the sequence (|h(yn)|p) is bounded and lim kynkp = ∞, we have a n→∞ contradiction.

Case 2. In this case the sequence (αn) is unbounded. Then we can find a ∞ strictly increasing sequence (nk)k=1 of positive integers such that

α1 ≤ β1 < αn1 ≤ βn1 < αn2 ≤ βn2 < ···≤ βnk−1 < αnk ≤ βnk .

∗ Define j0 = α1 and for k ≥ 1, set jk = αnk . Let h ∈ Ωp(N) be defined by

∞

h((zj)) = znk , (zj) ∈ Ωp(N). Xk=1 Then for all m ∈ N we have

∞

|h(ynm )|p = ynmnk = |ynmαnm |p = kynm kp. k=1 p X

Because the sequence (|h(yn )|p) is bounded and lim kyn kp = ∞, we again m m→∞ m have a contradiction. This completes the proof of the lemma.

Theorem 2.13. Let X ≃ Ωp(J), where J 6= ∅. Let A be a p-adic Banach algebra. If A ∈ A is X -analytic with compact spectrum σA 6= ∅, then A is A-analytic. Proof. We first prove that if A ∈ B(X ) is analytic with compact spectrum σA 6= ∅, then for all r> 0 there exists a constant Mr > 0 such that

kR(x; A)kp ≤ Mr for all x ∈ DσA (r).

Fix y ∈X and h ∈X ∗. Because A is analytic, Definition 1.14 implies that the function φ(x)= h(R(x; A)y), x ∈ σA, A CERTAIN p-ADICSPECTRALTHEOREM 61

is in H0(σA). Hence by Definition 1.10, lim φ(x)=0and φ(x) is the uniform |x|p→∞ limit on DσA (r) of rational functions with poles contained in σA. Let L > 0 P (x) be so large that |φ(x)| ≤ 1 for |x| ≥ L. Now let R(x)= be a rational p p Q(x) function with poles in σA such that

|φ(x) − R(x)|p ≤ 1 for all x ∈ DσA (r).

Now write n a0 + a1x + · · · + anx R(x)= n n , (x − b1) 1 · · · (x − bk) k where { b1, . . . , bnk }⊆ σA. Let x ∈ DσA (r), with |x|p ≤ L. Then for 1 ≤ j ≤ k, |x − bj|p ≥ r, because bj ∈ σA and dist(x,σA) ≥ r; hence we have 1 1 ≤ . nj nj |x − bj|p r Therefore we get n |a0 + a1x + · · · + anx |p |R(x)|p = n1 nk |x − b1|p ···|x − bk|p n max{ |a0|p, |a1|p|x|p,..., |an|p|x| } ≤ p rn1 · · · rnk max{ |a | , |a | L,..., |a | Ln } ≤ 0 p 1 p n p rn1 · · · rnk = M. It follows that

|φ(x) − R(x)|p + |R(x)|p ≤ 1+ M, if |x|p ≤ L and x ∈ DσA (r); |φ(x)|p ≤ (1, if |x|p ≥ L.

Therefore if we set Lr = M + 1 then |φ(x)|p ≤ Lr for all x ∈ DσA (r). It follows that the set

{ |h(R(x; A)y)|p | x ∈ DσA (r) } is bounded. Because h ∈X ∗ is arbitrary, Lemma 2.12 implies that the set

{ kR(x; A)ykp | x ∈ DσA (r) } is bounded. Hence, because y ∈ X is arbitrary, it follows from the Uniform Boundedness Theorem ([13],Theorem 3.12, p. 65) that the set

{ kR(x; A)kp | x ∈ DσA (r) } 62 R.L. Baker

is bounded above by some Mr > 0. Now let A be a p-adic Banach algebra and let A ∈ A be X -analytic with compact σA 6= ∅. Then by definition there exists an embedding θ : A→ B(X ) such that θ(A) is analytic in B(X )-analytic. Let r > 0. Then from what we proved above, there exists a constant Mr > 0 such that kR(x; θ(A))kp ≤ Mr −1 for all x ∈ Dσθ(A) (r). Let Nr = (Mr + 1)kθ kp. Then because σθ(A) ⊆ σA, it follows that DσA (r) ⊆ Dσθ(A) (r), and hence for all x ∈ DσA (r),

−1 kR(x; A)kp = kθ (R(x; θ(A)))kp −1 −1 ≤ kθ kpkR(x; θ(A))kp ≤ kθ kp(Mr +1) = Nr.

This completes the proof of the theorem.

Corollary 2.14. Let X ≃ Ωp(J), where J 6= ∅. Let A be a p-adic Banach algebra. Then for the X -analytic elements in A, Theorem 2.10 (the Extended p-adic Spectral Theorem) reduces to Theorem 1.21 (Vishik’s Spectral Theorem II).

Proof. Let A ∈ A have compact spectrum σA 6= ∅, and assume that A is X -analytic. Let f ∈ L(σA). Let 0 < r ∈ |Ωp|p be such that f ∈ Br(σA), where

N N − DσA (r)= Dai (r), σA ⊆ Dai (r ). (1) i i [=1 [=1 Here the union is assumed to be disjoint and each ai ∈ σA. Let Γ ∈ Ωp, with |Γ|p = r. By Theorem 2.13, A is A-analytic. By (f) of Lemma 1.8, L(σA) = L(σA) = F(σA), and hence we have f ∈ F(σA). It follows from (1) and Theorem 2.10 that

N f(A)= f(x)(x − ai)R(x; A) dx. i=1 Z Xai,Γ Hence Theorem 1.21 and Theorem 2.10 assign the same element of A to f(A).

By (g) of Lemma 1.8, we have Br,a1,...,aN = Br(σ), and hence statement (4) of Theorem 2.10 reduces to statement (3) of Theorem 1.21.

3. p-Adic UHF and TUHF Banach Algebras

Definition 3.1. A p-adic UHF algebra over Ωp is a unital p-adic Banach A CERTAIN p-ADICSPECTRALTHEOREM 63 algebra of the form ∞ A = Mn, n=1 [ where (Mn) is an increasing sequence of p-adic Banach subalgebras of A, such that each Mn contains the identity of A and is algebraically isomorphic as an

Ωp-algebra to the Ωp-algebra Mpn (Ωp) of pn × pn matrices over Ωp. A p-adic TUHF algebra over Ωp is a unital p-adic Banach algebra of the form

∞ T = Tn, n=1 [ where (Tn) is an increasing sequence of p-adic Banach subalgebras of T , such that each Tn contains the identity of T and is algebraically isomorphic as an

Ωp-algebra to the Ωp-algebra Tpn (Ωp) of pn × pn upper triangular matrices over Ωp. Here, for any positive integer m, the Ωp-algebra Mm(Ωp) of m × m m matrices over Ωp is identified with B(Ωp ) and Tm(Ωp) is the Ωp-algebra of m m × m upper triangular matrices in Mm(Ωp). Finally, Ωp is given the norm k(x1,...,xm)k = max |xi|p. 1≤i≤m

Lemma 3.2. Let X be a p-adic Banach space. with X ≃ Ωp(J), where J 6= ∅ is a finite set. Let A be a p-adic Banach algebra and assume that there exists a unital Ωp-monomorphism θ : A → B(X ). Then any A ∈ A is X -analytic. Proof. Let A ∈A. Because B(X ) is finite-dimensional, the results of finite- dimensional spectral theory apply to operators in B(X ). Therefore there exist x1,...,xN in Ωp such that σθ(A) = { x1,...,xN }. Moreover, there exist positive integers ν(x1),...,ν(xN ) and idempotents E(x1),...,E(xN ) in B(X ) such that for x ∈ σθ(A),

N ν(xj )−1 ν (θ(A) − xj) R(x; θ(A)) = ν+1 E(xj ). (1) (x − xj) j ν=0 X=1 X For a proof of (1), see [5], Theorem 10, page 560. Now fix y ∈X and h ∈X ∗. Then for x ∈ σθ(A), (1) implies that

N ν(xj )−1 hjν h(R(x; θ(A))y)= ν+1 , (2) (x − xj) j ν=0 X=1 X 64 R.L. Baker

ν where for 1 ≤ j ≤ N and 0 ≤ ν ≤ ν(xj) − 1, hjν = h((θ(A) − xj) E(xj)y). hjν Now, each of the functions ν+1 in (2) is clearly in H0(σθ(A)), hence (x − xj) (2) implies that the function Rh,y(x) = h(R(x; θ(A))y) belongs to H0(σθ(A)). Because y ∈X and h ∈X ∗ are arbitrary, we see that θ(A) is analytic. Because θ(A) is finite-dimensional, we see that θ−1 : θ(A) → A is bounded and θ(A) is a p-adic Banach subalgebra of B(X ) (see [13], Theorem 3.15, page 69). It follows that A is X -analytic. X -analytic.

Spectral 3.3. Theorem for p-adic UHF and TUHF Algebras. Let A be a p-adic UHF or a p-adic TUHF Banach algebra. Let A ∈A have spectrum σA.

Assume that for all r > 0, kR(x; A)kp is bounded on the complement DσA (r) − of DσA (r ). Then A is A-analytic. Let f, g ∈ F(σA). Let 0 < r ∈ |Ωp|p, and let Γ be in Ωp, with |Γ|p = r. Assume that a1, . . . , aN in Ωp are given, with

N − σA ⊆ Dai (r ), (1) i [=1 where the Dai (r) are disjoint. Finally, suppose that f and g are Krasner analytic on each Dai (r). By (1) and Definition 2.7, f(A) and g(A) are well-defined, with

N f(A)= f(x)(x − ai)R(x; A) dx, i=1 Z Xai,Γ N g(A)= g(x)(x − ai)R(x; A) dx. i=1 Z Xai,Γ

Let g ∈ F(A), and let α, β ∈ Ωp. Then αf + βg,f · g ∈ F(A), and

(2) (αf + βg)(A)= αf(A)+ βg(A);

(3) (f · g)(A)= f(A)g(A);

(4) The mapping h 7→ h(A) is continuous on Br,a1,...,aN .

Proof. A has the form ∞ A = An, n=1 [ A CERTAIN p-ADICSPECTRALTHEOREM 65

where for each n, there exists a finite nonempty set Jn and unital Ωp-monomor- phism θn from An into B(Xn), with Xn ≃ Ωp(Jn); because B(Xn) is finite- dimensional, it follows form Lemma 3.2 that the members of An are Xn-analytic. There exists a sequence (An) in

∞ An n=1 [ such that lim kA − Ankp = 0. Therefore A satisfies condition (b) of Definition n→∞ 1.19. By assumption, for all r > 0, kR(x; A)kp is bounded on the complement − DσA (r) of DσA (r ), and hence A satisfies condition (a) of Definition 1.19. It fol- lows that A is A-analytic. Properties (2)-(4) are then consequences of applying Theorem 2.10 to A. Theorem 3.4. Let A be a p-adic UHF or a p-adic TUHF Banach algebra. Let A ∈A have spectrum σA. Assume that for all r> 0, kR(x; A)kp is bounded − on the complement DσA (r) of DσA (r ). Then A is A-analytic. Let f ∈ F(A). Let 0 < r be in |Ωp|p, and let Γ be in Ωp, with |Γ|p = r. Assume that a1, . . . , aN in Ωp are given, with N − σA ⊆ Dai (r ), (1) i [=1 where the Dai (r) are disjoint and f is Krasner analytic on the Dai (r). Then for any ǫ > 0 there exists a δ > 0 such that if B ∈ A is A-analytic, with kA − Bkp < δ, then N

σB ⊆ Dai (r), i [=1 f ∈ F(B), and kf(A) − f(B)kp <ǫ. Proof. Because A and A satisfy the hypothesis of Theorem 3.3 it follows that A is A-analytic. Therefore the present lemma follows from an application of Lemma 2.8. Lemma 3.5. Let A be a p-adic Banach algebra and let E ∈ A be an idempotent. Then for all r > 0, kR(x; E)kp is bounded on DσE (r). If A is a p-adic UHF Banach algebra or a p-adic TUHF Banach algebra, then E satisfies the hypothesis of Theorem 3.3 and the hypothesis of Theorem 3.4.

Proof. Let r > 0 be arbitrary. Suppose that E = 0, then σE = {0}. If 1 x ∈ D (r), then |x| ≥ r, and hence kR(x; E)k = |x|−1 ≤ . This shows that σE p p p r 66 R.L. Baker

kR(x; E)kp is bounded on DσE (r). Now suppose that E = 1, then σE = {1}. 1 If x ∈ D (r), then |x − 1| ≥ r, and hence kR(x; E)k = |x − 1|−1 ≤ . We σE p p p r conclude that kR(x; E)kp is bounded on DσE (r). Now assume that E 6= 0, 1, then σE = {0, 1}. Let x ∈ Ωp, with |x|p > kEkp. Then by (a) of Theorem 2.1, E −1 E −1 1 − exists. Hence (x − E)−1 = x−1 1 − exists and x x     E −1 ∞ E n k(x − E)−1k = |x|−1 1 − = |x|−1 p p x p x   p n=0   p ∞ X kEk n ≤ |x|−1 p p |x| n p X=0   kEk −1 = |x|−1 1 − p . p |x|  p  2 It follows that lim kR(x; E)kp = 0. Let M ≥ 1, r , kEkp be so large that |x|p→∞

kR(x; E)kp ≤ 1 for |x|p ≥ M.

Let x ∈ DσE (r), with |x|p ≤ M. Then |x|p = |x − 0|p ≥ r and |x − 1|p ≥ r, hence we have 1 1 1 · ≤ 2 . |x|p |x − 1|p r Because E 6= 0, 1 it is easy to see that for x 6= 0, 1, (x − 1) + E R(x; E) = (x − E)−1 = . x(x − 1) Therefore we get

k(x − 1) + Ekp kR(x; E)kp = |x|p|x − 1|p max{|x − 1| , kEk } ≤ p p r2 max{1, |x| , kEk } ≤ p p r2 max{1, M, kEk } ≤ p r2 M = . r2 A CERTAIN p-ADICSPECTRALTHEOREM 67

M Thus, if we define N = , we see that r r2

kR(x; E)kp ≤ Nr for all x ∈ DσE (r).

Hence for all r> 0, kR(x; E)kp is bounded on DσE (r). Now assume that A is a p-adic UHF or TUHF algebra, and E ∈ A is an idempotent. A has the from ∞ A = An, n=1 [ where for each n, there exists a finite nonempty set Jn such that Xn ≃ Ωp(Jn) and an embedding θn : An → B(Xn); because B(Xn) is finite-dimensional, it follows form Lemma 3.2 that the members of An are Xn-analytic with nonempty spectrum. There exists a sequence (An) in ∞ An n [=1 such that lim kE − Ankp = 0. Now, we proved above that for all r > 0, n→∞ kR(x; E)kp is bounded on DσE (r), hence by Definition 1.19, E is A-analytic. This completes the proof of the theorem. As mentioned in the introduction of the present article, Theorem 3.3, The- orem 3.4, and Lemma 3.5 can be used to transfer, mutatis mutandis, the results of [2] to p-adic TUHF algebras, which is the content of [4]. The following key lemma is used in [4]; this lemma can be proved using Theorem 3.3, Theorem 3.4, and Lemma 3.5.

Lemma 3.6. Let A be a p-adic UHF or a p-adic TUHF algebra over Ωp. Let ǫ > 0 be positive number, and let I = { ei | 1 ≤ i ≤ n } be an orthogonal family of idempotents in A. Then there exists a positive number δ(ǫ,I) > 0 with the following property. Let B be a unital p-adic Banach subalgebra of A such that each member of B is A-analytic, and suppose that { ai | 1 ≤ i ≤ n } is a family of elements in B such that kei − aik ≤ δ(ǫ,I) for 1 ≤ i ≤ n. Then there exists an orthogonal family { fi | 1 ≤ i ≤ n } of idempotents in B such that kei − fik <ǫ for 1 ≤ i ≤ n.

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